Mathematics

# $\int { \dfrac { dx }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) } } =$

$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 6 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$

##### SOLUTION
Given,

$\int \dfrac{dx}{\left(x^2+1\right)\left(x^2+4\right)}$

$=\dfrac 13 \int \dfrac{3 \, dx}{\left(x^2+1\right)\left(x^2+4\right)}$

$=\dfrac 13 \int \dfrac{x^2+4-(x^2+1) \, }{\left(x^2+1\right)\left(x^2+4\right)}dx$

$=\int \dfrac{1}{3\left(x^2+1\right)}dx-\dfrac{1}{3\left(x^2+4\right)}dx$

$=\dfrac{1}{3}\tan ^{-1}\left(x\right)-\dfrac{1}{6}\tan ^{-1}\left(\dfrac{x}{2}\right)+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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